58 STEP 4. Review the Knowledge You Need to Score High
[–5,7] by [–40,20]
Figure 5.2-1Example 2
Find: limx→ 3 −
x^2
x^2 − 9.
Factor the denominator obtaining limx→ 3 −
x^2
x^2 − 9
=xlim→ 3 −
x^2
(x−3)(x+3). The limit of the numer-
ator is 9 and the limit of the denominator is (0)(6)=0 through negative values. Therefore,
xlim→ 3 −x^2
x^2 − 9=−∞. Verify your result with a calculator. (See Figure 5.2-2.)[–10,10] by [–10,10]
Figure 5.2-2Example 3Find: limx→ 5 −√
25 −x^2
x− 5.
Substituting 5 into the expression leads to 0/0. Factor the numerator√
√^25 −x^2 into
(5−x)(5+x). Asx → 5 −,(x −5) < 0. Rewrite (x −5) as−(5−x)asx →
5 −,(5−x) > 0 and thus, you may express (5−x)as√
(5−x)^2 =√
(5−x)(5−x).
Therefore, (x−5)=−(5−x)=−√
(5−x)(5−x). Substituting these equivalent expres-
sions into the original problem, you have limx→ 5 −√
25 −x^2
x− 5= xlim→ 5 −√
(5−x)(5+x)
√
(5−x)(5−x)=
−xlim→ 5 −√
(5−x)(5+x)
(5−x)(5−x)
=−xlim→ 5 −√
(5+x)
(5−x). The limit of the numerator is 10 and the limit
of the denominator is 0 through positive values. Thus, the limx→ 5 −√
25 −x^2
x− 5